$\frac{1}{2} x^2+3 x$
Answer (1)
Rewrite the expression as a function: f ( x ) = 2 1 x 2 + 3 x .
Find the x-coordinate of the vertex using the formula x v = − 2 a b , which gives x v = − 3 .
Find the y-coordinate of the vertex by substituting x v = − 3 into the function, resulting in f ( − 3 ) = − 2 9 .
Find the roots of the equation 2 1 x 2 + 3 x = 0 , which are x = 0 and x = − 6 . The vertex is ( − 3 , − 2 9 ) , the parabola opens upwards, and the roots are x = 0 and x = − 6 . f ( x ) = 2 1 x 2 + 3 x
Explanation
Understanding the Expression We are given the quadratic expression 2 1 x 2 + 3 x and we want to analyze it. This includes finding the vertex, roots, and determining the concavity of the parabola represented by this expression.
Rewriting as a Function Let's rewrite the expression as a function: f ( x ) = 2 1 x 2 + 3 x . This helps us analyze it more easily.
Finding the x-coordinate of the Vertex To find the vertex of the parabola, we use the formula for the x-coordinate of the vertex: x v = − 2 a b , where a = 2 1 and b = 3 . Plugging in these values, we get: x v = − 2 ( 2 1 ) 3 = − 1 3 = − 3.
Finding the y-coordinate of the Vertex Now, we find the y-coordinate of the vertex by substituting x v = − 3 into the function: f ( − 3 ) = 2 1 ( − 3 ) 2 + 3 ( − 3 ) = 2 1 ( 9 ) − 9 = 2 9 − 9 = 2 9 − 2 18 = − 2 9 = − 4.5. So, the vertex of the parabola is at ( − 3 , − 2 9 ) or ( − 3 , − 4.5 ) .
Determining the Concavity To determine the concavity of the parabola, we look at the coefficient of the x 2 term. Since 0"> a = 2 1 > 0 , the parabola opens upwards. This means the vertex is a minimum point.
Finding the Roots Now, let's find the roots of the quadratic equation 2 1 x 2 + 3 x = 0 . We can factor out an x :
x ( 2 1 x + 3 ) = 0.
Solving for the Roots Solving for x , we have two possibilities:
x = 0
2 1 x + 3 = 0 ⇒ 2 1 x = − 3 ⇒ x = − 6 So, the roots are x = 0 and x = − 6 .
Summary of Findings In summary, the vertex of the parabola is ( − 3 , − 2 9 ) , the parabola opens upwards, and the roots are x = 0 and x = − 6 .
Examples
Understanding quadratic expressions is crucial in various real-world applications. For instance, engineers use quadratic equations to model the trajectory of projectiles, such as designing the path of a rocket or the flight of a ball. Architects apply quadratic functions to design curved structures, like arches and bridges, ensuring stability and optimal use of materials. Economists also use quadratic functions to model cost and revenue curves, helping businesses determine the price point that maximizes profit. These applications demonstrate the practical significance of analyzing quadratic expressions.
